Question

There are 20 boys and 5 girls in a class. Three students are selected at random. If $E$ is an event of selecting one boy and two girls, then $P(E)$ is equal to

• A. $\frac{1}{23}$
• B. $\frac{4}{23}$
• C. $\frac{5}{23}$
• D. $\frac{3}{23}$
• E. $\frac{2}{23}$

Hint:

Always simplify your probability fractions at the end. Here, canceling the zeros makes the final fraction $\frac{2}{23}$ clear immediately.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
This is a classical probability problem involving combinations. We need to find the number of ways to select 1 boy and 2 girls, divided by the total number of ways to select any 3 students from the entire class.
Step 2: Key Formula or Approach:
The probability is given by \(P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}\).
Total students = 25.
Favorable = \(\binom{20}{1} \times \binom{5}{2}\).
Total = \(\binom{25}{3}\).
Step 3: Detailed Explanation:
Calculate the number of favorable ways:
Select 1 boy out of 20: \(\binom{20}{1} = 20\).
Select 2 girls out of 5: \(\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10\).
Total favorable ways = \(20 \times 10 = 200\).
Calculate the total number of ways to select 3 students from 25:
\[ \binom{25}{3} = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} \] \[ = 25 \times 4 \times 23 = 100 \times 23 = 2300 \] Calculate the probability:
\[ P(E) = \frac{200}{2300} = \frac{2}{23} \] Step 4: Final Answer:
The probability is \(\frac{2}{23}\).